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Theorem. There exists a nonempty set of rational numbers which is bounded above in $\mathbb{Q}$ but has no least upper bound in $\mathbb{Q}$.

Here is the relevant part of the proof that I am going to ask about.

Let $k=\frac{a}{b}\in\mathbb{Q}$ be an upper bound for the set $S=\{q\in\mathbb{Q}:q^2<2\}$. Suppose that $k^2<2$. Define $\delta=2-k^2>0$. Fix $N\in\mathbb{N}$ such that $$ N\geq\max\{2a+1,3a/(b^2\delta)\} $$ Since $N\geq 2a+1$, we have $N^2\geq N(2a+1)\geq 2Na+1$ and since $N>3a/(b^2\delta)$, then $N^2b^2\delta>3Na\geq 2Na+1$ which forces $$ \frac{2Na+1}{N^2b^2}<\delta $$ Define $l=\frac{Na+1}{Nb}$. Then $$ l^2=k^2+\frac{2Na+1}{N^2b^2}<k^2+\delta=2 $$

What is my question?

No matter how hard I think about this part of the proof I see nowhere that $N\geq 2a+1$ is used to show that $l^2<2$. I am sure that I am missing something quite obvious but I can not figure it out. Perhaps the thinking of the author of the proof was different to mine. Why was it necessary to fix $N\in\mathbb{N}$ such that $$ N\geq\max\{2a+1,3a/(b^2\delta)\} $$ and not just $N> 3a/(b^2\delta)$? In other words, I see $N\geq 2a+1$ unnecessary.

$k=\frac{a}{b}\in\mathbb{Q}$ is an upper bound in $\mathbb{Q}$ for $S$. We know that $\sqrt{2}\approx1.41421356237$ is the supremum of $S$ in real numbers so $k\geq 1.5$ or $k\geq 1.42$ or $k\geq 1.415$ and so on which forces $a>1$.

$l^2=k^2+\frac{2Na+1}{N^2b^2}$ and I can approximate $\frac{2Na+1}{N^2b^2}$ as follows

$$ \begin{align*} \frac{2Na+1}{N^2b^2}&\leq\frac{2Na+N}{N^2b^2}&&\text{since }N\geq 1\\ &=\frac{2a+1}{Nb^2}&&\\ &<\frac{2a+a}{Nb^2}&&\text{since }a>1\\ &=\frac{3a}{Nb^2}&& \end{align*} $$ So I need $\frac{3a}{Nb^2}<\delta$ which implies $N>\frac{3a}{b^2\delta}$.

  • It seems that this part of proof shows that no matter which rational you pick in the set, there is a bigger rational present in the set . So any upper bound of the set must not be in the set. The only part where $N\ge 2a+1$ is relevant here is in deducing $N^2\ge 2Na+1$ which implies $\frac{2Na+1}{N^2b^2}\le \frac{1}{b^2}$. Perhaps this fact would be used in later part of proof? – user160738 Apr 20 '17 at 02:50
  • @user160738: Thanks for looking into this. No, it is not used n later part of proof. –  Apr 20 '17 at 02:55
  • If $k$ is an upper bound with $k^2<2$, let $N>0$ so that $(k+1/N)^2=k^2+2k(1/N)+(1/N)^2<k^2+\delta=2$. You shouldn't have too much trouble showing such an $N$ exists. Then clearly $(k+1/N)^2<2$ and $k+1/N>k$. I'm not sure why your book is making this so complicated. – PVAL-inactive Apr 20 '17 at 02:59
  • @PVAL-inactive: I agree. In that case $N=\max(1,(1+2k)/\delta)$. –  Apr 20 '17 at 03:22
  • If I were to guess, I suspect the author was attempting to avoid numerically approximating $\sqrt{2}$ to get the bounds, and so used $N \geq 2a + 1$ to show: $3Na\geq 2Na+1$ - which perhaps the author would want to argue via $2Na + Na \geq 2Na + 1$, for which one would only need that $a$ is positive. – Artimis Fowl Apr 20 '17 at 06:10
  • @ArtimisFowl: This is my guess too but I fail to understand is that how $N\geq 2a+1$ is used to justify that $2Na+Na\geq 2Na+1$? $2Na+Na\geq 2Na+1$ is equivalent to $Na\geq 1$. If we multiply $N\geq 2a+1$ by $a$ (assuming $a>0$), then $Na\geq 2a^2+a$ so one yet has to show that $2a^2+a>1$. On the other hand, if this was the case, what was the need for $N^2\geq N(2a+1)\geq 2Na+1$? –  Apr 20 '17 at 06:24

1 Answers1

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The author presents a proof purely based upon calculations within $\mathbb{Q}$. It does not rely on an embedding of $\mathbb{Q}$ in $\mathbb{R}$ and the knowledge that $\mathbb{R}$ is complete.

Insofar is the approach of the author somewhat different than yours.

Let's revisit the author's proof. We want to show there does not exist a rational number $k$ which is a least upper bound of $S$, the set of rational numbers having square less than $2$.

In order to do so we show that whenever we want to take a least upper bound $k=\frac{a}{b}\in\mathbb{Q}$ of $S$ with $k^2<2$, we find a rational number $l$ in $S$ with $l>k$ contradicting the assumption $k$ is a least upper bound.

We consider the distance $\delta=2-k^2=2-\frac{a^2}{b^2}$ from the square of $k$ to $2$ and construct a rational number $l\in S$ having a smaller distance.

We take a natural number $N$ with \begin{align*} N\geq\max\{2a+1,3a/(b^2\delta)\} \end{align*}

Since $N\geq 2a+1$ and since $N>\frac{3a}{b^2\delta}=\frac{3a}{b^2\left(2-\frac{a^2}{b^2}\right)}=\frac{3a}{2b^2-a^2}$ we obtain \begin{align*} N^2b^2\delta&=N^2b^2\left(2-\frac{a^2}{b^2}\right)=N^2(2b^2-a^2)\\ &>3Na\\ &\geq 2Na+1 \end{align*} which implies \begin{align*} \frac{2Na+1}{N^2b^2}<\delta \end{align*}

If we now define $l:=\frac{Na+1}{Nb}>\frac{Na}{Nb}=k$ we obtain \begin{align*} l^2&=\frac{(Na+1)^2}{N^2b^2}=\frac{N^2a^2+2Na+1}{N^2b^2}\\ &=\frac{a^2}{b^2}+\frac{2Na+1}{N^2b^2}\\ &<\frac{a^2}{b^2}+\delta\\ &=2 \end{align*}

and the claim follows.

Note: In accordance with OP's comment below we don't need the following part of the author's proof: \begin{align*} N^2\geq N(2a+1)=2Na+N>2Na+1 \end{align*}

Markus Scheuer
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  • @Vafa: You rely on the existence of a supremum $\sqrt{2}\in \mathbb{R}\setminus\mathbb{Q}$, which the author doesn't. So, $a>1$ is not known per se. Is the author's proof clear to you without the assumption that $a>1$? – Markus Scheuer Apr 20 '17 at 08:15
  • The point of the proof is to show that $\mathbb{Q}$ is not complete, i.e. that there are numbers missing, and is a step towards defining its completion $\mathbb{R}$. You are using your knowledge of $\sqrt{2}\in \mathbb{R}$ in your version of the proof, which is kind of circular. The author's proof does not use knowledge of the reals. – Jaap Scherphuis Apr 20 '17 at 08:55
  • @MarkusScheuer: Thanks. So you are using $Na\geq N$. How does this inequality hold? –  Apr 20 '17 at 09:37
  • @Vafa: We may wlog assume $a\in\mathbb{N}$. – Markus Scheuer Apr 20 '17 at 09:39
  • @MarkusScheuer: Thanks a lot. That now makes complete sense. I have one last question. Why did we need $N^2\geq 2Na+1$? to me, it seems that $N\geq 2a+1$ was enough. –  Apr 20 '17 at 09:42
  • @Vafa: I agree. :-) Answer updated accordingly. – Markus Scheuer Apr 20 '17 at 10:02
  • @MarkusScheuer: Thanks a lot. If I could, I would upvote your answer more than one million times. I have no idea how to thank you for the time and effort that you spent to help me. –  Apr 20 '17 at 10:06
  • @Vafa: You're welcome! One upvote and accepting the answer is fine! Good to see the answer is useful. :-) – Markus Scheuer Apr 20 '17 at 10:08
  • @VahidDamanafshan: Thanks, Vahid! :-) – Markus Scheuer Apr 20 '17 at 13:56